Multiply the following complex numbers: $({3-2i}) \cdot ({5+i})$
Answer: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({3-2i}) \cdot ({5+i}) = $ $ ({3} \cdot {5}) + ({3} \cdot {1}i) + ({-2}i \cdot {5}) + ({-2}i \cdot {1}i) $ Then simplify the terms: $ (15) + (3i) + (-10i) + (-2 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 15 + (3 - 10)i - 2i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 15 + (3 - 10)i - (-2) $ The result is simplified: $ (15 + 2) + (-7i) = 17-7i $